Question

Volume of Co2, obtained at STP by the complete
decomposition of 9.85 gm Baco3is (Mol. wt. of
BaCO3 = 197)
(1) 2.24 litre
(2) 1.12 litre
(3) 0.85 litre
(4) 0.56 litre
H
eh...
​

Answer 1

Mass of Bariun Carbonate taken = 9.85 grams

Molar Mass of Barium Carbonate = 197 grams

Number of moles of BaCO3 taken will be :-

$moles=\dfrac{mass\:given}{molar\:mass}\\ \\ \\moles=\dfrac{9.85}{197}\\ \\ \\moles=0.05$

$\rule{200}{2}$

$\boxed{\sf{\bold{\Large{BaCO_3 \xrightarrow{HEAT} BaO+CO_2}}}}$

$\rule{200}{2}$

1 mole of BaCO3 produces 1 mole of CO2

Therefore,

0.05 mole of BaCO3 produces 0.05 mole of CO2

$\rule{200}{2}$

volume of CO2 produced at STP will be :-

$=moles*22.4\\ \\ \\=0.05*22.4\\ \\ \\=1.12\:L$

$\rule{200}{2}$

Hence,

option (2) is correct

$\rule{200}{2}$